=332 x (27 = 73) + 68 x ( 27 + 73)

= 332 x 100 + 68 x 100

= 33200 + 6800

= 4000

Còn câu b, thì hình như sai đề bạn ạ 

= 40.000 nha

Câu b không sai đề đâu bạn ạ!!!

a) (-25)x68+(-34)x(-250)

a) \(\left(x-\frac{1}{2}\right)^3=27\)

=> \(\left(x-\frac{1}{2}\right)^3=3^3\)

=> \(x-\frac{1}{2}=3\)

=> \(x=3+\frac{1}{2}\)

=> \(x=\frac{7}{2}\)

Vậy \(x=\frac{7}{2}.\)

b) \(\left(2x-1\right)^3=-27\)

=> \(\left(2x-1\right)^3=\left(-3\right)^3\)

=> \(2x-1=-3\)

=> \(2x=\left(-3\right)+1\)

=> \(2x=-2\)

=> \(x=\left(-2\right):2\)

=> \(x=-1\)

Vậy \(x=-1.\)

Chúc bạn học tốt!

a)\(\frac{x+32}{11}+\frac{x+23}{12}=\frac{x+38}{13}+\frac{x+27}{14}\)

\(\left(\frac{x-1}{11}+3\right)+\left(\frac{x-1}{12}+2\right)=\left(\frac{x-1}{13}+3\right)+\left(\frac{x-1}{14}+2\right)\)

\(\left(\frac{x-1}{11}+\frac{x-1}{12}\right)+\left(3+2\right)=\left(\frac{x-1}{13}+\frac{x-1}{14}\right)+\left(3+2\right)\)

\(\frac{x-1}{11}+\frac{x-1}{12}=\frac{x-1}{13}+\frac{x-1}{14}\)

\(\frac{x-1}{11}+\frac{x-1}{12}-\frac{x-1}{13}+\frac{x-1}{14}=0\)

\(\left(x-1\right)\left(\frac{1}{11}+\frac{1}{12}-\frac{1}{13}+\frac{1}{14}\right)=0\)

Vì \(\frac{1}{11}+\frac{1}{12}\ne\frac{1}{13}+\frac{1}{14}\)\(\Rightarrow\frac{1}{11}+\frac{1}{12}-\frac{1}{13}+\frac{1}{14}\ne0\)

\(\Rightarrow x-1=0\)

\(\Rightarrow x=1\)

cái này là lớp 7 mà

phần a ra x=1

kiến thức nâng cao à,

ax=1

b)x=2

12.x-27:24 = 28

12.x-\(\frac{9}{8}\)= 28

12x = 28+\(\frac{9}{8}\)

12x = \(29\frac{1}{8}\)

x = \(29\frac{1}{8}:12\)

x = \(\frac{233}{96}\)

(30+03+2004).2+x=(13+12+2004).2

2037.2+x=2009.2

4074+x=4058

x=4058-4074

x=-16

X= -16 PỢN Ạ

Lời giải:

a)

\(2006.2005^{2003}> 2005.2005^{2003}=2005^{1+2003}=2005^{2004}\)

Vậy \(2006.2005^{2003}> 2005^{2004}\)

b)

\(2005^{2004}+2005^{2003}=2005^{2003}(2005+1)=2005^{2003}.2006< 2006^{2003}.2006\)

hay \(2005^{2004}+2005^{2003}< 2006^{2004}\)

c) Thiếu đề

d)

\(72^{27}-72^{26}=72^{26}(72-1)=71.72^{26}\)

\(72^{28}-72^{27}=72^{27}(72-1)=71.72^{27}> 71.72^{26}\)

\(\Rightarrow 72^{28}-72^{27}> 72^{27}-72^{26}\)